ICSE Class 10 Physics Chapter 2 Exercise 2(A) Selina Solutions

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ICSE Class 10 Physics Selina Solutions – Chapter 2: Work, Energy and Power Exercise 2(A)

ICSE Class 10 Physics Chapter 2: Work, Energy and Power Exercise 2(A) is an important exercise for students preparing for the ICSE Class 10 Physics examination. This exercise covers fundamental concepts related to work, energy and power and includes conceptual as well as numerical-based questions.
These Selina Solutions for Class 10 Physics Chapter 2 Exercise 2(A) provide clear, step-by-step explanations to help students understand the questions and learn the correct method of solving them.

ICSE Class 10 Physics Chapter 2 Exercise 2(A) Solutions

The chapter Work, Energy and Power introduces students to the relationship between force, displacement, work and energy. Exercise 2(A) helps students strengthen their understanding of these basic concepts.
Students can use these ICSE Class 10 Physics Work, Energy and Power Exercise 2(A) Solutions for homework, revision and examination preparation.

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Study materials related to chapter 2: Work, Energy and Power

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☛ ICSE Class 10 Chapter 2 Work, Energy and Power Ex 2(B) Solutions

☛ ICSE Class 10 Chapter 2 Work, Energy and Power Ex 2(C) Solutions

☛ ICSE Class 10 Physics Chapter 2 – Work, Energy and Power Notes

☛ ICSE Class 10 Physics Chapter 2 – Work, Energy and Power Previous Year Questions

(A) Multiple Choice Type

(Choose the correct answer from the options given below).

Question 1
For work done to be maximum, the angle between the force and displacement should be :
(a) 0° (b) 30°
(c) 45° (d) 90°
Answer:
(a) 0°
Explanation:
Work done is given by the formula:
W = F × s × cos θ
When the displacement is in the same direction as the force, the angle between them is 0°.
Since cos 0° = 1, the work done becomes:
W = F × s × 1 = F × s
Hence, the work done is maximum when the force and displacement are in the same direction.
Therefore, the correct answer is (a) 0°.

Question 2
When a ball m is thrown upwards to a height h, the work done by the force of gravity is:
(a) mgh (b) – mgh
(c) mg (d) mgh²
Answer:
(b) – mgh
Explanation:
Gravity acts downward while the ball moves upward. Since force and displacement are in opposite directions (θ = 180°), W = f × s × cos 180° = – mgh

Question 3
Identify the incorrect statement :
No work is done by a force F on an object if :
(a) the object is stationary but the point of application of force moves on the object.
(b) the object moves in such a way that the point of application of force remains fixed.
(c) the force is always perpendicular to velocity.
(d) the force is always perpendicular to acceleration.
Answer:
(d) the force is always perpendicular to acceleration.
Explanation:
Work depends on the angle between force and displacement, not between force and acceleration.

(a) Correct – If the point of application does not undergo displacement in the direction of force, work is zero.
(b) Correct – If the point of application remains fixed, no work is done.
(c) Correct – If force is perpendicular to velocity (and hence displacement), work done is zero.
(d) Incorrect – A force cannot generally remain perpendicular to its own acceleration because acceleration is produced by force. This condition does not determine whether work is done.

Question 4
A coolie A takes 1 minute to lift a load to the roof of a bus, whereas coolie B take 2 minutes to lift the same load to the roof of the same bus. The work done by B is:
(a) equal to work done by A
(b) greater than work done by A
(c) less than work done by A
(d) none of the above
Answer:
(a) equal to work done by A
Explanation:
Both coolies lift the same load through the same height.
Work done = mgh

Question 5
The energy of electron is 2.5 eV. When expressed in Joule, it is:
(a) 1.6 x 10^–19 J (b) 4.0 x 10^–19 J
(c) 3.2 x 10^–19 J (d) 0
Answer:
(b) 4.0 x 10^–19 J
Explanation:
Given,
energy of electron = 2.5 eV
1 eV = 1.6 x 10^–19 J
Hence,
2.5 eV = 2.5 x 1.6 x 10^–19 = 4 x 10^–19 J

Question 6
A block of mass 20 kg is pulled up a slope as shown in the figure given below with a constant speed by applying a force of 200 N parallel to the slope from the initial position A to the final position B. Work done in moving the block from A to B is:

(a) 600 J (b) 500 J
(c) 60 J (d) 200 J
Answer:
(a) 600 J
Explanation:
Given:

Force, F = 200 N
Displacement along the slope, s = 3 m

Using the formula:
Work done = Force × Displacement
⇒ W = F × s
⇒ W = 200 × 3
⇒ W = 600 J
Therefore, the work done in moving the block from A to B is 600 J.

Question 7
The graph given below shows a force-displacement graph. What is the total work done by force on the body in displacement from x = 0 to x = 6 m ?

(a) 20 J (b) 40 J
(c) 60 J (d) 80 J
Answer:
(b) 40 J
Explanation:
The work done is equal to the area under the force–displacement graph.
1. From x = 0 to 2 m (Triangle):
Area = $\frac{1}{2}$ × Base × Height
= $\frac{1}{2}$ × 2 × 10
= 10 J
2. From x = 2 to 4 m (Rectangle):
Area = Length × Breadth
= 2 × 10
= 20 J
3. From x = 4 to 6 m (Triangle):
Area = $\frac{1}{2}$ × Base × Height
= $\frac{1}{2}$ × 2 × 10
= 10 J
Total work done:
W = 10 + 20 + 10
W = 40 J
Therefore, the correct answer is (b) 40 J.

Question 8
Assertion (A): When the displacement is normal to the direction of force, the work done is zero.
Reason (R): Work done depends on the angle between force and displacement.
(a) both A and R are true and R is the correct explanation of A
(b) both A and R are true but R is not the correct explanation of A
(c) assertion is false but reason is true
(d) assertion is true but reason is false
Answer:
(a) Both A and R are true and R is the correct explanation of A.
Explanation:
Assertion (A) is true.
Work done is given by:
W = F × d × cos θ
Where:

F = magnitude of force
d = displacement
θ = angle between the force and the displacement

If the displacement is perpendicular (90°) to the direction of the force, then:
cos 90° = 0
Therefore,
W = F × d × 0 = 0
Hence, the work done is zero.
Reason (R) is also true because the work done depends on the angle between the force and the displacement. Since the angle is 90°, the work done becomes zero.
Therefore, both Assertion (A) and Reason (R) are true, and Reason (R) correctly explains Assertion (A).

Question 9
Assertion (A): Work done by frictional force is negative.
Reason (R): This is because frictional force acts along the direction of motion.
(a) Both A and R are true and R is the correct explanation of A
(b) Both A and R are true and R is not the correct explanation of A
(c) Assertion is false but reason is true
(d) Assertion is true but reason is false
Answer:
(d) Assertion is true but reason is false.
Explanation:
Friction acts opposite to the direction of motion, not along it. Since the angle between frictional force and displacement is 180°,
W = f × s × 180° = – f × s

(B) Very Short Questions

Question 1
A force F acts on a body and displaces it by a distance S in a direction at an angle θ with the direction of force.
(a) Write the expression for the work done by the force.
(b) What should be the angle between force and displacement so that the work done is (i) zero, (ii) maximum?
Answer:
(a) When a force acts at an angle θ to the direction of displacement, the work done is given by:
W = F × S × cos θ
(b) (i) Work done is zero:
The work done is zero when the angle between the force and the displacement is 90°, because cos 90° = 0.
Therefore,
W = F × S × cos 90° = 0
(ii) Work done is maximum:
The work done is maximum when the angle between the force and the displacement is 0°, because cos 0° = 1.
Therefore,
W = F × S × cos 0° = F × S

Question 2
A body is moved in a direction opposite to the direction of force acting on it. State whether the work is done by the force or work is done against the force.
Answer:
Work is done against the force.

Question 3
State whether work is done or not by writing yes or no, in the following cases?
(a) A man pushes a wall.
(b) A coolie stands with a box on his head for 15 min.
(c) A boy climbs up 20 stairs.
Answer:
(a) No, work is not done by the man.
(b) No, work is not done by the coolie.
(c) Yes, work is done by the boy in climbing up the 20 stairs.

Question 4
Give an example when work done by the force of gravity acting on a body is zero even though the body gets displaced from its initial position.
Answer:
A person carrying a load while walking on a horizontal road.

Question 5
A boy of mass m climbs up a stairs of vertical height h.
(a) What is the work done by the boy against the force of gravity?
(b) What would have been the work done if he uses a lift in climbing the same vertical height?
Answer:
(a) Work done against gravity = mgh
(b) Work done = mgh (same, because the vertical height is the same)

Question 6
What physical quantity does the electron volt (eV) measure? How is it related to the S.I. unit of that quantity?
Answer:
The energy of atomic particles is measured in electron volt (eV).
1 eV = 1.6 x 10^–19 J

Question 7
Complete the following sentences :
(a) 1 J = …………… calorie.
(b) 1 kWh = …………… J.
Answer:
(a) 1 J = 0.24 calorie.
(b) 1 kWh = 3.6 x 10⁶ J.

Question 8
Name the physical quantity which is measured in calorie. How is it related to the S.I unit of that quantity?
Answer:
The physical quantity which is measured in calorie is heat energy.
1 calorie = 4.2 joule

Question 9
(a) Name the physical quantity measured in terms of horsepower.
(b) How is horsepower related to the S. I. unit of power?
Answer:
(a) Power
(b) 1 horsepower (hp) = 746 W

Question 10
Name the quantity which is measured in
(a) kWh (b) kW (c) Wh (d) eV
Answer:
(a) kWh – Energy
(b) kW – Power
(c) Wh – Energy
(d) eV – Energy

Question 11
Is it possible that no transfer of energy takes place even when a force is applied to a body?
Answer:
Yes. If there is no displacement, no energy is transferred and no work is done.

(C) Short Questions

Question 1
Define work. When is work said to be done by a force?
Answer:
Work is said to be done when a force acting on a body produces displacement in the direction of the force.
Work is done only when a force applied on a body produces displacement.

Question 2
How is the work done by a force measured when
(i) force is in the direction of displacement,
(ii) force is at an angle to the direction of displacement?
Answer:
(i) If force is in the direction of displacement:
W = F × S
(ii) If force makes an angle θ with displacement:
W = F × S × cos θ

Question 3
A body is acted upon by a force. State two conditions when the work done is zero.
Answer:

When there is no displacement.
When the force acts perpendicular (90°) to the displacement.

Question 4
When a body moves in a circular path, how much work is done by the body? Give reason.
Answer:
Work done is zero because the centripetal force is always perpendicular to the direction of motion.

Question 5
A satellite revolves around the earth in a circular orbit. What is the work done by the force of gravity on the satellite? Give reason.
Answer:
The work done by gravity on the satellite is zero because gravity acts perpendicular to the satellite’s displacement.

Question 6
A coolie X carrying a load on his head climbs up a slope and another coolie Y carrying the identical load on his head move the same distance on a frictionless horizontal platform. Who does more work? Explain the reason.
Answer:
Coolie X does more work because he gains height and works against gravity. Coolie Y moves horizontally on a frictionless surface, so no work is done against gravity.

Question 7
The work done by a fielder when he takes a catch in a cricket match is negative. Explain.
Answer:
While taking a catch, the fielder applies a force opposite to the motion of the ball. Therefore, the work done by the fielder is negative.

Question 8
What are the S.I. and C.G.S. units of work? How are they related? Establish the relationship.
Answer:

S.I. unit: Joule (J)
C.G.S. unit: Erg
Relation: 1 J = 10⁷ erg

Question 9
State and define the S.I. unit of work.
Answer:
The S.I. unit of work is joule (J).
One joule is the work done when a force of 1 N moves a body through 1 m in the direction of the force.

Question 10
Express joule in terms of erg.
Answer:
1 joule = 1 N × 1 m
1 joule = 10⁵ dyne × 10² cm [∵ 1 N = 10⁵ dyne, 1 m = 10² cm]
1 joule = 10⁷ dyne × cm
𝟏 𝐣𝐨𝐮𝐥𝐞 = 𝟏𝟎^𝟕 𝐞𝐫𝐠

Question 11
Define the term energy and state its S.I unit.
Answer:
Energy is the capacity to do work.
S.I. unit: Joule (J)

Question 12
Define kilowatt hour. How is it related to joule?
Answer:
A kilowatt-hour (kWh) is the energy consumed by a 1 kW appliance in 1 hour.
1 kWh = 3.6 × 10⁶ J

Question 13
Define the term power. State its S.I. unit.
Answer:
Power is defined as the rate of doing work. The S.I unit of power is watt (W)

Question 14
Differentiate between work and power.
Answer:

Work	Power
Work is the product of the applied force and the displacement produced in its direction.	Power is the amount of work done per unit time.
Work done does not depend on the time taken.	Power depends on the time taken to perform the work.
The S.I. unit of work is joule (J).	The S.I. unit of power is watt (W).

Question 15
Differentiate between energy and power.
Answer:

Energy	Power
Energy is the capacity to do work.	Power is the rate of doing work.
Energy spent is independent of time.	Power depends on the time in which energy is spent.
S.I. unit: Joule (J)	S.I. unit: Watt (W)

Question 16
State and define the S.I unit of power.
Answer:
The S.I. unit of power is watt (W).
One watt is the power when 1 joule of work is done in 1 second.
1 W = 1 Js^–1

Question 17
Differentiate between watt and watt-hour.
Answer:

Watt (W)	Watt-hour (Wh)
Unit of power	Unit of energy
Measures the rate of doing work	Measures the amount of energy consumed

(D) Long Questions

Question 1
State the condition when the work done by a force is (a) positive, (b) negative. Explain with the help of examples.
Answer:
(a) Positive Work Done: Work done is positive when the force acts in the same direction as the displacement of the body.
Examples:

A person pushing a trolley forward.
A freely falling stone under gravity.

(b) Negative Work Done: Work done is negative when the force acts opposite to the direction of displacement.
Examples:

Friction acting on a moving body.
Gravity acting on a body moving vertically upward.

Question 2
A body of mass m falls through a height h. Obtain an expression for the work done by the force of gravity.
Answer:
Force due to gravity = mg
Displacement = h
Since force and displacement are in the same direction,
W = F × s
W = mg × h
Therefore,
W = mgh
Hence, the work done by gravity is positive and is equal to mgh.

Question 3
State two factors on which power spent by a source depends. Explain your answer with examples.
Answer:
Power is the rate of doing work.
$\mathrm{P}=\frac{\mathrm{W}}{\mathrm{t}}$

Power depends on:

Work done: Greater the work done in the same time, greater is the power.
Example: A crane lifting a heavier load develops more power.
Time taken: For the same work, less the time taken, greater is the power.
Example: Two persons lift the same load to the same height. The one who takes less time develops more power.

(E) Numericals

Question 1
A body, when acted upon by a force of 20 kgf, gets displaced by 1 m. Calculate the work done by the force, when the displacement is:
(i) in the direction of force,
(ii) at an angle of 60° with the force, and
(iii) normal to the force. (g = 10 N kg^–1)
Solution:
Given,

Force, F = 20 kgf = 20 × 10 N = 200 N
Displacement, S = 1 m

(i) Along the direction of force:
Work done = Force × Displacement
⇒ W = F × S
⇒ W = 200 × 1
⇒ W = 200 J
Therefore, work done is 200 J.

(ii) At 60°,
W = F × S cos θ
⇒ W = 200 × 1 × cos 60°
⇒ W = 200 × $\frac{1}{2}$
⇒ W = 100 J
Therefore, work done is 100 J.

(iii) Normal to the force:
W = F × S cos θ
⇒ W = 200 × 1 × cos 90°
⇒ W = 200 × 1 × 0
⇒ W = 0 J
Therefore, work done is 0 J.

Question 2
A boy of mass 40 kg climbs up the stairs and reaches the roof at a height 8 m in 5 s. Calculate:
(i) the force of gravity acting on the boy,
(ii) the work done by him against the force of gravity,
(iii) the power spent by the boy. (Take g = 10 m s^–2)
Solution:
Given,

Mass (m) = 40 kg
Height (h) = 8 m
Time (t) = 5 s
g = 10 m/s²

(i) Force of gravity:
⇒ F = mg
⇒ F = 40 × 10
⇒ F = 400 N
Hence, the force of gravity acting on the boy is 400 N.

(ii) Work done:
⇒ W = mgh
⇒ W = 40 × 10 × 8
⇒ W = 3200 J
Hence, the work done by the boy against the force of gravity is 3200 J.

(iii) Power, $\mathrm{P}=\ \frac{\mathrm{W}}{\mathrm{t}}$
⇒ $\mathrm{P}=\ \frac{3200}{5}$
⇒ P = 640 W
Hence, power spent by the boy is 640 W.

Question 3
A man spends 7.4 kJ energy in displacing a body by 74 m in the direction in which he applies force, in 2.5 s.
Calculate:
(i) the force applied, and
(ii) the power spent (in H.P) by the man.
Solution:
Given,

Energy (W) = 7.4 kJ = 7.4 × 1000 J = 7400 J
Displacement (S) = 74 m
Time (t) = 2.5 s

(i) Force, $\mathrm{F}\ =\ \frac{\mathrm{W}}{\mathrm{s}}$
⇒ $\mathrm{F}\ =\ \frac{7400}{74}$
⇒ F = 100 N
Hence, the force applied is 100 N.

(ii) Power spent $=\frac{\mathrm{Work}\ \mathrm{done}}{\mathrm{Time}\ \mathrm{taken}}$
⇒ P $=\frac{7400}{2.5}$
⇒ P = 2960 W
We know,
1 HP = 746 W
Power ≈ 3.97 HP
Hence, the power spent by the man is 3.97 H.P.

Question 4
A weight lifter lifted a load of 200 kgf to a height of 2.5 m in 5 s.
Calculate:
(i) the work done, and
(ii) the power developed by him.
Take g = 10 N kg^–1
Solution:
Given,

Load (F) = 200 kgf = 200 × 10 N = 2000 N
Height (h) = 2.5 m
Time (T) = 5 s

(i) Work done:
⇒ W = Force × Height
⇒ W = 2000 × 2.5
⇒ W = 5000 J
⇒ Therefore, work done is 5000 J.

(ii) Power $=\frac{\mathrm{Work}\ \mathrm{done}}{\mathrm{Time}\ \mathrm{taken}}$
⇒ P = $\frac{5000}{5}$
⇒ P = 1000 W
⇒ P = 1 kW
Therefore, power developed by the boy is 1000 W (1 kW).

Question 5
An electric heater of power 3 kW is used for 10 h. How much energy does it consume?
Express your answer in:
(i) kWh,
(ii) joule.
Solution:
Given,

Power (P) = 3 kW
Time (t) = 10 h

(i) Energy consumed in kWh:
Energy = Power × Time
Energy = 3 × 10
Energy = 30 kWh
So, energy consumed is 30 kWh.

(ii) Energy consumed in joule:
1 kWh = 3.6 × 10⁶ J
Energy = 30 × 3.6 × 10⁶
W = 1.08 × 10⁸ J
So, energy consumed is 1.08 x 10⁸ J.

Question 6
A pump is used to lift 600 kg of water from a depth of 75 m in 10 s.
Calculate:
(a) the work done by the pump,
(b) the power at which the pump works, and
(c) the power rating of the pump if its efficiency is 40 %. (Take g = 10 m s^–2).
Solution:
Given,

Mass of water, m = 600 kg
Height, h = 75 m
Time taken, t = 10 s
Acceleration due to gravity, g = 10 m/s²
Efficiency = 40%

(a) Work done by the pump:
We know,
Work done = mgh
Substituting the given values,
⇒ W = 600 × 10 × 75
⇒ W = 450000 J
⇒ W = 4.5 × 10⁵ J
Therefore, the work done by the pump is 4.5 × 10⁵ J.

(b) Power of the pump:
We know,
Power $=\frac{\mathrm{Work}\ \mathrm{done}\ }{\mathrm{Time}\ \mathrm{taken}}$
P $=\frac{45000}{10}$ = 45000 W = 45 kW
Therefore, the power developed by the pump is 45 kW.

(c) Power rating of the pump:
We know,
Efficiency $=\frac{\mathrm{Useful}\ \mathrm{power}\ }{\mathrm{Input}\ \mathrm{power}}$
Given,
Efficiency = 40% = $\frac{40}{100}$ = 0.4
Therefore,
0.4 $=\frac{45}{\mathrm{Input}\ \mathrm{power}}$
Input power $=\frac{45}{0.4}=$ 112.5 kW
Therefore, the power rating of the pump is 112.5 kW.

Question 7
An ox can apply a maximum force of 1000 N. It is taking part in a cart race and is able to pull the cart at a constant speed of 30 m s^–1 while making its best effort. Calculate the power developed by the ox.
Solution:
Given,

Force = 1000 N
Speed = 30 m/s

Power = Force × Speed
⇒ P = 1000 × 30
⇒ P = 30000 W
⇒ P kW
⇒ P = 30 kW

Question 8
The power of a motor is 40 kW. At what speed can the motor raise a load of 20,000 N?
Solution:
Given,

Power = 40 kW = 40000 W
Load = 20000 N

⇒ Speed $=\frac{\mathrm{Power}}{\mathrm{Force}}$
⇒ Speed $=\frac{40000}{20000}$
⇒ Speed = 2 m/s

Question 9
Rajan exerts a force of 150 N in pulling a cart at a constant speed of 10 m s^–1. Calculate the power exerted.
Solution:
Given,

Force = 150 N
Speed = 10 m/s

Power = Force × Speed
⇒ Power = 150 × 10
⇒ Power = 1500 W

Question 10
A boy weighing 350 N climbs up 30 steps, each 20 cm high in 1 minute.
Calculate:
(i) the work done, and
(ii) the power spent.
Solution:

(i) We know,
Work done = Force × Distance moved in the direction of force
W = F × s
Given,
● Force, F = 350 N
● Number of steps = 30
● Height of each step = 20 cm = 0.2 m
Total distance covered,
s = 30 × 0.2 = 6 m
Substituting the values,
W = 350 × 6
W = 2100 J
Therefore, the work done by the boy is 2100 J.

(ii) Power Developed:
We know,
Power $=\frac{\mathrm{Work}\ \mathrm{done}}{\mathrm{Time}\ \mathrm{taken}}$
Time taken = 1 minute = 60 s
P = $\frac{2100}{60}$ = 35 W
Therefore, the power developed by the boy is 35 W.

Question 11
It takes 20 s for a person A of mass 50 kg to climb up the stairs, while another person B of same mass does the same in 15 s.
Compare the:
(i) work done, and
(ii) power developed by the persons A and B.
Solution:
(i) We know that work done is the product of force and displacement and is independent of time.
Since both persons have the same mass and climb the same height, they do the same amount of work.
Therefore,
Work done by A : Work done by B = 1 : 1
(ii) We know,
Power $=\frac{\mathrm{Work}\ \mathrm{done}\ }{\mathrm{Time}\ \mathrm{taken}}$
Since both persons do the same amount of work,
Power developed by A : Power developed by B
= $\frac{\mathrm{W}}{20}\ :\ \frac{\mathrm{W}}{15}$
= 15 : 20
= 3 : 4
Therefore, the ratio of the power developed by Person A to that developed by Person B is 3 : 4.

Question 12
A boy weighing 40 kgf climbs up a stair of 30 steps, each 20 cm high in 4 minutes and a girl weighing 30 kgf does the same in 3 minutes.
Compare:
(i) the work done by them, and
(ii) the power developed by them.
Solution:

(i) We know,
Work done = Force × Displacement = mgh
Given,

Weight of boy (m₁) = 40 kgf
Weight of girl (m₂) = 30 kgf
Time taken by boy (t₁) = 4 min = 240 s
Time taken by girl (t₂) = 3 min = 180 s
Height climbed (h) = 30 × 20 cm = 600 cm = 6 m

Since both climb the same height,
Work done by boy : Work done by girl
= m₁gh : m₂gh
= m₁ : m₂
= 40 : 30
= 4 : 3
Therefore, the ratio of work done by the boy to that done by the girl is 4 : 3.

(ii) We know,
Power $=\frac{\mathrm{Work}\ \mathrm{done}\ }{\mathrm{Time}\ \mathrm{taken}}$
Power developed by boy : Power developed by girl = 1 : 1
$=\frac{m_{1}gh}{t_{1}}\ :\ \frac{m_{2}gh}{t_{2}}$
$=\frac{m_{1}}{t_{1}}\ :\ \frac{m_{2}}{t_{2}}$
$=\frac{40}{240}\ :\ \frac{30}{180}$
$=\frac{1}{6}\ :\ \frac{1}{6}$ = = 1 : 1
Therefore, the ratio of the power developed by the boy to that developed by the girl is 1 : 1.

Question 13
A man raises a box of mass 50 kg to a height of 2 m in 20 s, while another man raises the same box to the same height in 50 s.
(a) Compare : (i) the work done, and (ii) the power developed by them.
(b) Calculate : (i) the work done, and (ii) the power developed by each man.
Take g = 10 N kg^–1.
Solution:
Given,

Mass = 50 kg
Height = 2 m
g = 10 m/s²

(a) (i) Work done is the same for both men.
Therefore, Work done by first man : Work done by second man = 1 : 1
(ii) Let the work done by each man be W.
Power is given by:
Power $=\frac{\mathrm{Work}\ \mathrm{done}\ }{\mathrm{Time}\ \mathrm{taken}}$
Therefore,
Power developed by first man : Power developed by second Man
= $\frac{\mathrm{W}}{20}\ :\ \frac{\mathrm{W}}{50}$
= 50 : 20
= 5 : 2

(b) (i) Work done by each man:
We know,
W = F × s = mgh
W = 50 × 10 × 2
W = 1000 J
Therefore, the work done by each man is 1000 J.

(ii) Power developed:
For First Man:
$\mathrm{P}=\ \frac{\mathrm{W}}{\mathrm{t}}$
$\Rightarrow \ \mathrm{P}=\ \frac{1000}{20}$
$⇒$ P = 50 W
For Second Man:
$\mathrm{P}=\ \frac{\mathrm{W}}{\mathrm{t}}$
$\Rightarrow \ \mathrm{P}=\ \frac{1000}{50}$
⇒ P = 20 W
Therefore, the power developed by First Man is 50 W and the power developed by Second Man is 20 W.

Question 14
A boy takes 3 minutes to lift a 20 litre water bucket from a 20 m deep well, while his father does it in 2 minutes.
(a) Compare:
(i) the work, and (ii) power developed by them.
(b) How much work each does?
Take density of water = 10³ kg m^–3 and g = 9.8 N kg^–1
Solution:
Given,

Volume = 20 litres = 20/1000 m³ = 0.02 m³
Density = 1000 kg/m³
Height = 20 m
g = 9.8 m/s²
Mass = 1000 × 0.02 = 20 kg

(a) (i) Work done is the same for both.
So,
Work done by boy : Work done by father = 1 : 1
(ii) Comparing the Power Developed:
Power is given by:
Power $=\frac{\mathrm{Work}\ \mathrm{done}\ }{\mathrm{Time}\ \mathrm{taken}}$
Since both the boy and his father do the same amount of work,
Power of boy : Power of father
= $\frac{\mathrm{W}}{3}\ :\ \frac{\mathrm{W}}{2}$
= 2 : 3
Therefore, Power developed by boy : Power developed by father = 2 : 3

(b) Work done:
W = mgh
⇒ W = 20 × 9.8 × 20
⇒ W = 3920 J
∴ Work done by each = 3920 J

Question 15
A body of mass 5 kg moves along a straight line under the action of a variable force. From x = 0 to x = 4 m, the force acts in the direction of displacement and increases linearly from 0 N to 20 N. From x = 4 m to x = 6 m, the motion of the body reverses while the force acts in forward direction. From x = 6 m to x = 10 m, the force becomes negative acting opposite to the original direction of motion. Calculate the net work done by the force on the body.
Solution:

Work done equals the area under the force-displacement graph.

Region I: x = 0 to 4 m
The graph forms a triangle with:
● Base = 4 m
● Height = 20 N
Therefore,
W₁ = $\frac{1}{2}$ × base × height
= $\frac{1}{2}$ × 4 × 20
= + 40 J

Region II: x = 4 to 6 m
The area is a triangle with:
● Base = 2 m
● Height = 20 N
Therefore,
W₂ = $\frac{1}{2}$ × (− 2) × 20
= − 20 J

Region III: x = 6 to 10 m
● Base = 10 − 6 = 4 m
● Height = 20 N
Therefore,
W₃ = $\frac{1}{2}$ × (− 4) × (− 20)
= + 40 J

Net work done:
W = W₁ + W₂ + W₃
W = 40 − 20 + 40
W = 60 J
Hence, the net work done by the force on the body is 60 J.

Download Selina Solutions for ICSE Class 10 Physics Exercise 2(A) – Work, Energy and Power

You can download free PDF solutions for Selina Class 10 Physics Exercise 2(A) to revise offline. These solutions are exam-ready and designed by subject experts.

Why Practise Selina Exercise 2(A)?

The Selina Concise Physics textbook is widely used by ICSE Class 10 students for Physics preparation. Solving the exercises helps students understand concepts and develop numerical problem-solving skills.

Regular practice of ICSE Class 10 Physics Chapter 2 Exercise 2(A) Solutions can help students:

Understand important Physics concepts.

Learn the correct formulas.
Improve numerical-solving skills.
Revise the chapter effectively.
Prepare for examination questions.
Identify common mistakes.
Build confidence in Physics.

ICSE Class 10 Physics Chapter 2 Revision Tips

For effective preparation of Work, Energy and Power, students should:

Read the theory carefully.
Understand every definition.
Learn all important formulas.
Practise numerical problems regularly.
Pay attention to SI units.
Solve the textbook exercises.
Practise previous year questions.
Revise difficult questions before the examination.

Conclusion

ICSE Class 10 Physics Chapter 2: Work, Energy and Power Exercise 2(A) is an important exercise for understanding the basic principles of work and energy. Students should practise the questions carefully and understand the formulas instead of simply memorising the answers.